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Domain and Range Problem

How would I find the domain of
f(x)=x2+1,x≥0f\left(x\right)=x^2+1, x\ge 0
Inverse was  f−1(x)=x−1\:f^{-1}\left(x\right)=\sqrt{x-1}

Without drawing a graph.
Original post by joyoustele
How would I find the domain of
f(x)=x2+1,x≥0f\left(x\right)=x^2+1, x\ge 0
Inverse was  f−1(x)=x−1\:f^{-1}\left(x\right)=\sqrt{x-1}

Without drawing a graph.


You already specified its domain.
x>=0 is the domain lol
Original post by joyoustele
How would I find the domain of
f(x)=x2+1,x≥0f\left(x\right)=x^2+1, x\ge 0
Inverse was  f−1(x)=x−1\:f^{-1}\left(x\right)=\sqrt{x-1}

Without drawing a graph.


You're already given the domain of f(x) right there in the question: x >= 0, so presumably you're asking about the domain of f^-1(x). The answer is that since the inverse function maps outputs of f(x) back to the inputs that produced them, the domain of the inverse function is the set of possible outputs of f(x), i.e. the range of f(x). Here, the domain of f(x) is x >= 0, so x^2 can be any non-negative number, and thus x^2 + 1 can be any number > = 1. Thus the range of f(x) is f(x) >= 1, so the domain of f^-1(x) is x >= 1.
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joyoustele
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Original post by RDKGames
You already specified its domain.


No, the domain for the inverse is what I want to know how to do
Original post by joyoustele
No, the domain for the inverse is what I want to know how to do


Domain of the inverse is the range of the original. The original is half of a parabola.
Domain of inverse = range of f(x)
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Avatar for joyoustele
joyoustele
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19
Original post by Prasiortle
You're already given the domain of f(x) right there in the question: x >= 0, so presumably you're asking about the domain of f^-1(x). The answer is that since the inverse function maps outputs of f(x) back to the inputs that produced them, the domain of the inverse function is the set of possible outputs of f(x), i.e. the range of f(x). Here, the domain of f(x) is x >= 0, so x^2 can be any non-negative number, and thus x^2 + 1 can be any number > = 1. Thus the range of f(x) is f(x) >= 1, so the domain of f^-1(x) is x >= 1.


Thanks, easier than I thought

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